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I want to prove this exercise:

Let $a_n \to a$ and $b_n \to b$ for $n \to \infty$ Prove that, $a_n \cdot b_n \to a \cdot b$.

My proof:

Let $\epsilon > 0$ and $N_1 \in \mathbb{N}$ such that $|a_n - a| < \epsilon \forall n \geq N_1 $ and $N_2 \in \mathbb{N}$ such that $|b_n - b| < \epsilon \forall n \geq N_2 $ Then:

$$|a_nb_n - ab| = |a_nb_n - a_nb + a_nb - ab|=|a_nb_n - ab| \leq |a_nb_n - ab| \leq |\epsilon - \epsilon| = 0$$

Are theses conclusions correct?

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    $\begingroup$ For a start, you can't just choose $\varepsilon$ to be equal to some value. You need to let $\varepsilon > 0$ be arbitrary. $\endgroup$ – lokodiz Aug 29 '13 at 14:46
  • $\begingroup$ @SimonC changed it... $\endgroup$ – user2051347 Aug 29 '13 at 14:52
  • $\begingroup$ The inequality $|a_nb_n - ab| \leq |\epsilon - \epsilon|$ is false. $\endgroup$ – Umberto P. Aug 29 '13 at 14:53
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    $\begingroup$ Notice that the third term in your equality is the same as you first term. As is the fourth term. $\endgroup$ – lokodiz Aug 29 '13 at 14:54
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Some general hints:

  • $|a_n b_n - ab| \le |a_n b_n - a_n b| + |a_n b - ab|$ (why?)
  • You can factorise the above
  • Since $a_n \rightarrow a$, $|a_n|$ can be bounded by some $M > 0$

I can elaborate more, but try to make use of these.

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  • $\begingroup$ Thx for your answer! I do not get it... Please elaborate more! $\endgroup$ – user2051347 Aug 30 '13 at 20:49
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The calculations look a bit redundant and in the last step you can't have $|\epsilon - \epsilon|$ for $|a_n b_n - ab|$. Also as a comment pointed out, you need to consider arbitrary $\epsilon > 0$ and show $|a_n b_n - ab| < \epsilon$ for sufficiently large $n$. A hint on how to proceed: If $a_n = a + \delta_n$ and $b_n = b + \gamma_n$, then $a_n b_n = ab + \delta_n b + \gamma_n a + \delta_n \gamma_n$. For given $\epsilon > 0$, you just need that $|\gamma_n a| < \epsilon / 3$, and $|\delta_n b| < \epsilon / 3$, and $|\delta_n \gamma_n| < \epsilon / 3$.

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